Sum of all the prime divisors of a number

Difficulty Level : Medium
Last Updated : 24 Jun, 2021

Given a number N. The task is to find the sum of all the prime divisors of N.

Examples:

Input: 60
Output: 10
2, 3, 5 are prime divisors of 60

Input: 39
Output: 16
3, 13 are prime divisors of 39

A naive approach will be to iterate for all numbers till N and check if the number divides N. If the number divides N, check if that number is prime or not. Add all the prime numbers till N which divides N.

Below is the implementation of the above approach:

C++

// CPP program to find sum of
// prime divisors of N
#include <bits/stdc++.h>
using namespace std;
#define N 1000005
 
// Function to check if the
// number is prime or not.
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// function to find sum of prime
// divisors of N
int SumOfPrimeDivisors(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            if (isPrime(i))
                sum += i;
        }
    }
    return sum;
}
// Driver code
int main()
{
    int n = 60;
    cout << "Sum of prime divisors of 60 is " << SumOfPrimeDivisors(n) << endl;
}

Java

// Java program to find sum
// of prime divisors of N
import java.io.*;
import java.util.*;
 
class GFG
{
// Function to check if the
// number is prime or not.
static boolean isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that
    // we can skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5;
             i * i <= n; i = i + 6)
        if (n % i == 0 ||
            n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// function to find
// sum of prime
// divisors of N
static int SumOfPrimeDivisors(int n)
{
    int sum = 0;
    for (int i = 1;
             i <= n; i++)
    {
        if (n % i == 0)
        {
            if (isPrime(i))
                sum += i;
        }
    }
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int n = 60;
    System.out.print("Sum of prime divisors of 60 is " +
                          SumOfPrimeDivisors(n) + "\n");
}
}

C#

// C# program to find sum
// of prime divisors of N
using System;
class GFG
{
     
// Function to check if the
// number is prime or not.
static bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that
    // we can skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5;
             i * i <= n; i = i + 6)
        if (n % i == 0 ||
            n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// function to find
// sum of prime
// divisors of N
static int SumOfPrimeDivisors(int n)
{
    int sum = 0;
    for (int i = 1;
            i <= n; i++)
    {
        if (n % i == 0)
        {
            if (isPrime(i))
                sum += i;
        }
    }
    return sum;
}
 
// Driver code
public static void Main()
{
    int n = 60;
    Console.WriteLine("Sum of prime divisors of 60 is " +
                        SumOfPrimeDivisors(n) + "\n");
}
}
 
// This code is contributed
// by inder_verma.

Python3

# Python 3 program to find
# sum of prime divisors of N
N = 1000005
 
# Function to check if the
# number is prime or not.
def isPrime(n):
     
    # Corner cases
    if n <= 1:
        return False
    if n <= 3:
        return True
 
    # This is checked so that 
    # we can skip middle five
    # numbers in below loop
    if n % 2 == 0 or n % 3 == 0:
        return False
 
    i = 5
    while i * i <= n:
        if (n % i == 0 or
            n % (i + 2) == 0):
            return False
        i = i + 6
 
    return True
 
# function to find sum
# of prime divisors of N
def SumOfPrimeDivisors(n):
    sum = 0
    for i in range(1, n + 1) :
        if n % i == 0 :
            if isPrime(i):
                sum += i
     
    return sum
 
# Driver code
n = 60
print("Sum of prime divisors of 60 is " +
              str(SumOfPrimeDivisors(n)))
 
# This code is contributed
# by ChitraNayal

PHP

<?php
// PHP program to find sum
// of prime divisors of N
$N = 1000005;
 
// Function to check if the
// number is prime or not.
function isPrime($n)
{
    global $N;
    // Corner cases
    if ($n <= 1)
        return false;
    if ($n <= 3)
        return true;
 
    // This is checked so that
    // we can skip middle five
    // numbers in below loop
    if ($n % 2 == 0 || $n % 3 == 0)
        return false;
 
    for ($i = 5; $i * $i <= $n;
                 $i = $i + 6)
        if ($n % $i == 0 ||
            $n % ($i + 2) == 0)
            return false;
 
    return true;
}
 
// function to find sum
// of prime divisors of N
function SumOfPrimeDivisors($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
    {
        if ($n % $i == 0)
        {
            if (isPrime($i))
                $sum += $i;
        }
    }
    return $sum;
}
 
// Driver code
$n = 60;
echo "Sum of prime divisors of 60 is " .
                 SumOfPrimeDivisors($n);
 
// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// Javascript program to find sum of
// prime divisors of N
    let N = 1000005;
     
    // Function to check if the
    // number is prime or not.
    function isPrime(n)
    {
        // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
   
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
   
    for (let i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
   
    return true;
    }
     
    // function to find sum of prime
    // divisors of N
    function SumOfPrimeDivisors(n)
    {
        let sum = 0;
    for (let i = 1; i <= n; i++) {
        if (n % i == 0) {
            if (isPrime(i))
                sum += i;
        }
    }
    return sum;
    }
     
    // Driver code
    let n = 60;
    document.write("Sum of prime divisors of 60 is "+SumOfPrimeDivisors(n));
     
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output

Sum of prime divisors of 60 is 10

Time Complexity: O(N * sqrt(N))

Efficient Approach: The complexity can be reduced using Sieve of Eratosthenes with some modifications. The modifications are as follows:

Take an array of size N and substitute zero in all the indexes(initially consider all the numbers are prime).
Iterate for all the numbers whose indexes have zero(i.e., it is prime numbers).
Add this number to all it’s multiples less than N
Return the array[N] value which has the sum stored in it.

Below is the implementation of the above approach.

C++

// CPP program to find prime divisors of
// all numbers from 1 to n
#include <bits/stdc++.h>
using namespace std;
 
// function to find prime divisors of
// all numbers from 1 to n
int Sum(int N)
{
    int SumOfPrimeDivisors[N+1] = { 0 };
 
    for (int i = 2; i <= N; ++i) {
 
        // if the number is prime
        if (!SumOfPrimeDivisors[i]) {
 
            // add this prime to all it's multiples
            for (int j = i; j <= N; j += i) {
 
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
    return SumOfPrimeDivisors[N];
}
 
// Driver code
int main()
{
    int N = 60;
    cout << "Sum of prime divisors of 60 is "
         << Sum(N) << endl;
}

Java

// Java program to find
// prime divisors of
// all numbers from 1 to n
import java.io.*;
import java.util.*;
 
class GFG
{
     
// function to find prime
// divisors of all numbers
// from 1 to n
static int Sum(int N)
{
    int SumOfPrimeDivisors[] = new int[N + 1];
     
 
    for (int i = 2; i <= N; ++i)
    {
 
        // if the number is prime
        if (SumOfPrimeDivisors[i] == 0)
        {
 
            // add this prime to
            // all it's multiples
            for (int j = i; j <= N; j += i)
            {
 
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
    return SumOfPrimeDivisors[N];
}
 
// Driver code
public static void main(String args[])
{
    int N = 60;
    System.out.print("Sum of prime " +
                "divisors of 60 is " +
                       Sum(N) + "\n");
}
}

Python3

# Python 3 program to find
# prime divisors of
# all numbers from 1 to n
 
# function to find prime
# divisors of all numbers
# from 1 to n
def Sum(N):
  
    SumOfPrimeDivisors = [0] * (N + 1)
      
    for i in range(2, N + 1) :
      
        # if the number is prime
        if (SumOfPrimeDivisors[i] == 0) :
          
            # add this prime to
            # all it's multiples
            for j in range(i, N + 1, i) :
              
                SumOfPrimeDivisors[j] += i
              
    return SumOfPrimeDivisors[N]
  
# Driver code
N = 60
print("Sum of prime" ,
      "divisors of 60 is",
                  Sum(N));
                   
# This code is contributed
# by Smitha

C#

// C# program to find
// prime divisors of
// all numbers from 1 to n
using System;
 
class GFG
{
     
// function to find prime
// divisors of all numbers
// from 1 to n
static int Sum(int N)
{
    int []SumOfPrimeDivisors = new int[N + 1];
     
    for (int i = 2; i <= N; ++i)
    {
 
        // if the number is prime
        if (SumOfPrimeDivisors[i] == 0)
        {
 
            // add this prime to
            // all it's multiples
            for (int j = i;
                     j <= N; j += i)
            {
 
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
     
    return SumOfPrimeDivisors[N];
}
 
// Driver code
public static void Main()
{
    int N = 60;
    Console.Write("Sum of prime " +
                    "divisors of 60 is " +
                         Sum(N) + "\n");
}
}
 
// This code is contributed
// by Smitha

PHP

<?php
// PHP program to find prime
// divisors of all numbers
// from 1 to n
 
// function to find prime
// divisors of all numbers
// from 1 to n
function Sum($N)
{
    for($i = 0; $i <= $N; $i++)
        $SumOfPrimeDivisors[$i] = 0;
 
    for ($i = 2; $i <= $N; ++$i)
    {
 
        // if the number is prime
        if (!$SumOfPrimeDivisors[$i])
        {
 
            // add this prime to
            // all it's multiples
            for ($j = $i; $j <= $N; $j += $i)
            {
 
                $SumOfPrimeDivisors[$j] += $i;
            }
        }
    }
    return $SumOfPrimeDivisors[$N];
}
 
// Driver code
$N = 60;
echo "Sum of prime divisors of 60 is " . Sum($N);
 
// This code is contributed by Mahadev99
?>

Javascript

<script>
 
// Javascript program to find
// prime divisors of
// all numbers from 1 to n
     
    // function to find prime
    // divisors of all numbers
    // from 1 to n
    function Sum(N)
    {
        let SumOfPrimeDivisors = new Array(N+1);
        for(let i=0;i<SumOfPrimeDivisors.length;i++)
        {
            SumOfPrimeDivisors[i]=0;
        }
        for (let i = 2; i <= N; ++i)
        {
  
            // if the number is prime
            if (SumOfPrimeDivisors[i] == 0)
            {
  
                // add this prime to
                // all it's multiples
                for (let j = i; j <= N; j += i)
                {
  
                    SumOfPrimeDivisors[j] += i;
                }
            }
        }
        return SumOfPrimeDivisors[N];
    }
     
    // Driver code
    let N = 60;
    document.write("Sum of prime " +
                "divisors of 60 is " +
                       Sum(N) + "<br>");
     
    // This code is contributed by rag2127
     
</script>

Output

Sum of prime divisors of 60 is 10

Time Complexity: O(N * log N)

Efficient Approach:

Time complexity can be reduced by finding all the factors efficiently.

Below approach describe how to find all the factors efficiently.

If we look carefully, all the divisors are present in pairs. For example if n = 100, then the various pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)

Using this fact we could speed up our program significantly.

We, however, have to be careful if there are two equal divisors as in the case of (10, 10). In such case, we’d take only one of them.

Below is the implementation of the above approach.

C++

// C++ program to find sum of
// prime divisors of N
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the
// number is prime or not.
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// function to find sum of prime
// divisors of N
int SumOfPrimeDivisors(int n)
{
    int sum = 0;
    // return type of sqrt function
    // if float
    int root_n = (int)sqrt(n);
    for (int i = 1; i <= root_n; i++) {
        if (n % i == 0) {
            // both factors are same
            if (i == n / i && isPrime(i)) {
                sum += i;
            }
            else {
                // both factors are
                // not same ( i and n/i )
                if (isPrime(i)) {
                    sum += i;
                }
                if (isPrime(n / i)) {
                    sum += (n / i);
                }
            }
        }
    }
    return sum;
}
// Driver code
int main()
{
    int n = 60;
    cout << "Sum of prime divisors of 60 is "
         << SumOfPrimeDivisors(n) << endl;
}
// This code is contributed by hemantraj712

Java

// Java program to find sum of
// prime divisors of N
class GFG{
 
// Function to check if the
// number is prime or not.
static boolean isPrime(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
   
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
   
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
   
    return true;
}
   
// Function to find sum of prime
// divisors of N
static int SumOfPrimeDivisors(int n)
{
    int sum = 0;
     
    // Return type of sqrt function
    // if float
    int root_n = (int)Math.sqrt(n);
    for(int i = 1; i <= root_n; i++)
    {
        if (n % i == 0)
        {
             
            // Both factors are same
            if (i == n / i && isPrime(i))
            {
                sum += i;
            }
            else
            {
                 
                // Both factors are
                // not same ( i and n/i )
                if (isPrime(i))
                {
                    sum += i;
                }
                if (isPrime(n / i))
                {
                    sum += (n / i);
                }
            }
        }
    }
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 60;
    System.out.println("Sum of prime divisors of 60 is " +
                       SumOfPrimeDivisors(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find sum of
# prime divisors of N
import math
 
# Function to check if the
# number is prime or not.
def isPrime(n) :
 
    # Corner cases
    if (n <= 1) :
        return False
    if (n <= 3) :
        return True
  
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0) :
        return False
     
    i = 5
    while i * i <= n :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
  
    return True
  
# function to find sum of prime
# divisors of N
def SumOfPrimeDivisors(n) :
 
    Sum = 0
     
    # return type of sqrt function
    # if float
    root_n = (int)(math.sqrt(n))
    for i in range(1, root_n + 1) :
        if (n % i == 0) :
           
            # both factors are same
            if (i == (int)(n / i) and isPrime(i)) :
                Sum += i
             
            else :
                # both factors are
                # not same ( i and n/i )
                if (isPrime(i)) :
                    Sum += i
                 
                if (isPrime((int)(n / i))) :
                    Sum += (int)(n / i)
                    
    return Sum
     
n = 60
print("Sum of prime divisors of 60 is", SumOfPrimeDivisors(n))
 
# This code is contributed by rameshtravel07

C#

// C# program to find sum of
// prime divisors of N
using System;
class GFG {
     
    // Function to check if the
    // number is prime or not.
    static bool isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
      
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
      
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
      
        return true;
    }
      
    // function to find sum of prime
    // divisors of N
    static int SumOfPrimeDivisors(int n)
    {
        int sum = 0;
        // return type of sqrt function
        // if float
        int root_n = (int)Math.Sqrt(n);
        for (int i = 1; i <= root_n; i++) {
            if (n % i == 0) {
                // both factors are same
                if (i == n / i && isPrime(i)) {
                    sum += i;
                }
                else {
                    // both factors are
                    // not same ( i and n/i )
                    if (isPrime(i)) {
                        sum += i;
                    }
                    if (isPrime(n / i)) {
                        sum += (n / i);
                    }
                }
            }
        }
        return sum;
    }
 
  static void Main() {
    int n = 60;
    Console.WriteLine("Sum of prime divisors of 60 is " + SumOfPrimeDivisors(n));
  }
}
 
// This code is contributed by suresh07.

Javascript

<script>
    // Javascript program to find sum of
    // prime divisors of N
     
    // Function to check if the
    // number is prime or not.
    function isPrime(n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (let i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // function to find sum of prime
    // divisors of N
    function SumOfPrimeDivisors(n)
    {
        let sum = 0;
        // return type of sqrt function
        // if float
        let root_n = parseInt(Math.sqrt(n), 10);
        for (let i = 1; i <= root_n; i++) {
            if (n % i == 0) {
                // both factors are same
                if (i == parseInt(n / i, 10) && isPrime(i)) {
                    sum += i;
                }
                else {
                    // both factors are
                    // not same ( i and n/i )
                    if (isPrime(i)) {
                        sum += i;
                    }
                    if (isPrime(parseInt(n / i, 10))) {
                        sum += (parseInt(n / i, 10));
                    }
                }
            }
        }
        return sum;
    }
     
    let n = 60;
    document.write("Sum of prime divisors of 60 is "
         + SumOfPrimeDivisors(n) + "</br>");
  
 // This code is contributed by muksh07.
</script>

Output

Sum of prime divisors of 60 is 10

Time Complexity: O(sqrt(N) * sqrt(N))

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